Question

A tire manufacturer believes that the treadlife of its snow tires can be described by a Normal model with a mean of 32,000 miles and standard deviation of 2500 miles. a) If you buy a set of these tires, would it be reasonable for you to hope they'll last 40,000 miles? Explain. b) Approximately what fraction of these tires can be expected to last less than 30,000 miles? c) Approximately what fraction of these tires can be expected to last between 30,000 and 35,000 miles? d) Estimate the IQR of the treadlives. e) In planning a marketing strategy, a local tire dealer wants to offer a refund to any customer whose tires fail to last a certain number of miles. However, the dealer does not want to take too big a risk. If the dealer is willing to give refunds to no more than 1 of every 25 customers, for what mileage can he guarantee these tires to last?

Short answer

a) No, 40,000 miles is unreasonable. b) 21.19%. c) 67.3%. d) 3,375 miles. e) 27,625 miles for refund guarantee.

Step-by-step solution

Understand the Distribution

The treadlife of the tires is modeled by a normal distribution with mean \( \mu = 32,000 \) miles and standard deviation \( \sigma = 2,500 \) miles.

Assess 40,000 Mile Hope

Calculate the \( z \)-score for 40,000 miles using the formula \( z = \frac{x - \mu}{\sigma} \) where \( x = 40,000 \). This results in \( z = \frac{40,000 - 32,000}{2,500} = 3.2 \). A \( z \)-score of 3.2 corresponds to a very small probability (approximately 0.07%) of occurrence, making it unreasonable to expect the tires to last 40,000 miles.

Calculate Fraction Less Than 30,000 Miles

Calculate the \( z \)-score for 30,000 miles: \( z = \frac{30,000 - 32,000}{2,500} = -0.8 \). Using the standard normal distribution table or calculator, the probability for \( z = -0.8 \) is approximately 0.2119, meaning about 21.19% of tires last less than 30,000 miles.

Calculate Fraction Between 30,000 and 35,000 Miles

Find \( z \)-scores for 30,000 miles (\( z = -0.8 \)) and 35,000 miles (\( z = \frac{35,000 - 32,000}{2,500} = 1.2 \)). The probability for \( z = 1.2 \) is approximately 0.8849. Thus, the fraction of tires lasting between 30,000 and 35,000 miles is \( 0.8849 - 0.2119 = 0.673 \), or 67.3%.

Estimate the IQR of Treadlives

The first quartile (Q1) has a \( z \)-score of -0.675 and the third quartile (Q3) a \( z \)-score of 0.675. Convert these \( z \)-scores back to mileage: \( Q1 = 32,000 + (-0.675 \times 2,500) = 30,312.5 \) and \( Q3 = 32,000 + (0.675 \times 2,500) = 33,687.5 \). IQR is \( Q3 - Q1 = 33,687.5 - 30,312.5 = 3,375 \) miles.

Refund Mileage for 1 in 25 Customers

For 1 in 25 customers, the dealer should refund for performance below the 4th percentile. The corresponding \( z \)-score for the 4th percentile is approximately -1.75. Convert this \( z \)-score to a mileage: \( x = 32,000 + (-1.75 \times 2,500) = 27,625 \) miles. Therefore, the refund should apply to tires lasting less than approximately 27,625 miles.

Key concepts

z-score

A z-score, also known as a standard score, is a numerical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. For instance, in the tire treadlife problem, we calculate a z-score to understand how unusual or typical a treadlife value is when compared to the distribution mean of 32,000 miles with a standard deviation of 2,500 miles.
The formula for z-score is given by: \( z = \frac{x - \mu}{\sigma} \)

• \( x \) is the value of interest (mileage in this context).
• \( \mu \) is the mean of the distribution.
• \( \sigma \) is the standard deviation.

For example, to calculate how likely it is for a tire to last 40,000 miles, we compute its z-score: \( z = \frac{40,000 - 32,000}{2,500} = 3.2 \)
A z-score of 3.2 means that 40,000 miles is 3.2 standard deviations above the mean, which implies a very low probability (about 0.07%) that a randomly selected tire would last that long.

probability

Probability in the context of the normal distribution helps us understand the likelihood of certain outcomes. It is essential when assessing the probability of tire treadlife exceeding or falling between specified mileage benchmarks.
For example, when we want to determine the probability of tires lasting less than 30,000 miles, we use the z-score of -0.8 (calculated earlier) and consult the standard normal distribution table which shows that approximately 21.19% of tires can be expected to last less than 30,000 miles.
Additionally, to find the probability of tires lasting between 30,000 and 35,000 miles, we use two z-scores: -0.8 for 30,000 miles and 1.2 for 35,000 miles. The probability of lasting up to 35,000 miles is about 88.49%, while up to 30,000 miles is about 21.19%. Thus, the probability of falling within this interval is the difference: 67.3%.

standard deviation

The standard deviation is a measure of the amount of variation or dispersion of a set of values. In the normal distribution, it provides insights into how spread out the values are around the mean. For the tire treadlife, a standard deviation of 2,500 miles tells us how much the treadlife of different tires might vary from the average of 32,000 miles.
It helps in:

• Calculating z-scores for various mileages.
• Estimating probabilities of different treadlife mileages.

A smaller standard deviation indicates that the data points are closer to the mean, while a larger standard deviation means they are more spread out. In our case, the standard deviation helps dealers understand the range of mileages they can expect a typical tire to last.

quartiles

Quartiles divide a data set into four equal parts, offering a way to understand the spread and central tendency of the data. The first quartile (Q1) is the middle number between the smallest number and the median of the data set, while the third quartile (Q3) is the middle value between the median and the highest value.
In the treadlife problem, finding the interquartile range (IQR) helps tire dealers understand the middle 50% of the tire's lifespan. The IQR is calculated by finding the difference between Q3 and Q1:

• Q1 corresponds to a z-score of -0.675, \( Q1 = 32,000 + (-0.675 \times 2,500) = 30,312.5 \)
• Q3 corresponds to a z-score of 0.675, \( Q3 = 32,000 + (0.675 \times 2,500) = 33,687.5 \)

The IQR is \( 33,687.5 - 30,312.5 = 3,375 \) miles. It provides a useful summary of the spread of the central 50% of the data and can be critical in making informed decisions.